Find the sum: 1+111+11111+1111111+…..1….111(2k+1) ones
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Can you somehow manipulate the expression in geometric progression?
\(\frac{1}{9}(9+99+999+9999.........)\)
Now, can u transfer it into G.P series?
Ok let's see,
$\frac{1}{9}\left[\left(10^{1}-1\right)+\left(10^{2}-1\right)+\ldots \ldots\right]$ upto 2k+1 terms
= $\frac{1}{9}\left[\left(10+10^{2}+10^{3}+\ldots \ldots\right)\\-(1+1+1+1 \ldots \ldots . .)\right]$
Now can u see the G.P Series, with first term 10 and common ratio 10?
Ok let's see,
$\frac{1}{9}\left[\left(10^{1}-1\right)+\left(10^{2}-1\right)+\ldots \ldots \right]$ upto 2k+1 terms
= $\frac{1}{9}\left[\left(10+10^{2}+10^{3}+\ldots \ldots\right)\\-(1+1+1+1 \ldots \ldots . .)\right]$
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